Question

he number of square feet per house are normally distributed with an unknown population mean and standard deviation. If a random sample of 47 houses is taken to estimate the mean house size, what t-score should be used to find a 80% confidence interval estimate for the population mean?

Answer 1

Answer: 1.300228

Explanation:

We know that the t-score for a level of confidence
`(1-\alpha)` is given by :_

`t_((df,\alpha/2))`, whre df is the degree of freedom and
`\alpha` is the significance level.

Given : Level of significance :
`1-\alpha:0.80`

Then , significance level :
`\alpha: 1-0.80=0.20`

Since , sample size :
`n=47`

Degree of freedom for t-distribution:
`df=n-1=47-1=46`

With the help of the normal t-distribution table, we have

`t_((df,\alpha/2))=t_(46,0.10)=1.300228`

Hence, the t-score should be used to find a 80% confidence interval estimate for the population mean = 1.300228